Question

$\left(i\right)$ Differentiate the given function w.r.t. $x.$
$\sqrt{3x+2}+\frac{1}{\sqrt{2x^2+4}}$

$\left(ii\right)$ Differentiate the given function w.r.t. $x.$
$e^{{\sec }^{2}x+3{\cos }^{-1}x}$

$\left(iii\right)$ Differentiate the given function w.r.t. $x$
${\text{log}}_7\left(\text{log}x\right)$

Answer 1

$\left(i\right)$ Let $y=\sqrt{3x+2}+\frac{1}{\sqrt{2x^2+4}}$
Differentiating w.r.t. $x$ , we get,

$\frac{dy}{dx}=\frac{d(\sqrt{3x+2}+\frac{1}{\sqrt{2x^2+4}})}{dx}$

$\frac{dy}{dx}=\frac{d\left(\sqrt{3x+2}\right)}{dx}+\frac{d\left(\frac{1}{\sqrt{2x^2+4}}\right)}{dx}$

Using chain rule : $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx}=\frac{1}{2\sqrt{3x+2}}\times \frac{d(3x+2)}{dx}-\frac{1}{2}(2x^2+4{)}^{\frac{-1}{2}-1}.\frac{d(2x^2+4)}{dx}$

$\frac{dy}{dx}=\frac{1}{2\sqrt{3x+2}}\times (3+0)-\frac{1}{2}(2x^2+4{)}^{\frac{-3}{2}}.(4x+0)$

$\frac{dy}{dx}=\frac{3}{2\sqrt{3x+2}}-\frac{2x}{(2x^2+4{)}^{\frac{3}{2}}}$

$\left(ii\right)$ Let $y=e^{{\sec }^{2}x+3{\cos }^{-1}x}$
Differentiating w.r.t. $x$ , we get,

$\frac{dy}{dx}=\frac{e^{{\sec }^{2}x+3{\cos }^{-1}x}}{dx}$

$\frac{dy}{dx}=\frac{d(e^{{\sec }^{2}x)}}{dx}+\frac{d\left(3{\cos }^{-1}x\right)}{dx}$

Using chain rule : $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx}=e^{{\sec }^{2}x}.2\sec x.\frac{d\left(\sec x\right)}{dx}-\frac{3}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=e^{{\sec }^{2}x}.2\sec x.\sec x\cdot \tan x-\frac{3}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=2e^{{\sec }^{2}x}.{\sec }^{2}x.\tan x-\frac{3}{\sqrt{1-x^2}}$

$\left(iii\right)$ Let $y={\text{log}}_7\left(\text{log}x\right)$

$y=\frac{\text{log}\left(\text{log}x\right)}{{\text{log}}_7}$

[ logarithmic property ]
(Differentiation of $\text{log}x$ base e is known)

Differentiating w.r.t. $x$ , we get,

$\frac{dy}{dx}=\frac{1}{\text{log}7}\times \frac{d\left(\text{log}\right(\text{log}x\left)\right)}{dx}$

Using chain rule : $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx}=\frac{1}{\text{log}7}.\frac{1}{\text{log}x}\frac{d\left(\text{log}x\right)}{dx}$

$\frac{dy}{dx}=\frac{1}{\text{log}7}.\frac{1}{\text{log}x}\frac{1}{x}$

$\frac{dy}{dx}=\frac{1}{x\text{log}7\text{log}x}$